# Integrals involving trig functions

1. Jul 17, 2010

### Chandasouk

1. The problem statement, all variables and given/known data

I need help evaluating the integral Cotx3/10

I factored out the 1/10 from the integral and am just left with (1/10)*Cotx3

from here i do not really know what to do. I rewrote it in terms of sine and cosine to get

(1/10)*(Cosx3/Sinx3)dx

I multiply the integral by (1/Sinx3) to get rid of the denominator and am left with

(1/10)*(Cosx3dx

I factor out a (Cosx2) and am left with

(1/10)*(Cosx2)(Cosx)dx

Rewriting using trig indentities, I get

(1/10)*(1-Sinx2)(Cosx)dx

rewriting i get

(1/10)*(cosx-sinx2cosx)dx

I solve this integral and get

(1/10)sinx - (1/10)*((sinx)3/3) + C

but that is incorrect. The answer is supposed to be

(-1/20)cotx2-(1/10)ln(sinx)+C

What went wrong?

2. Jul 17, 2010

### vela

Staff Emeritus
Well, for one thing, you can't just get rid of parts of the integrand.
Just to be clear, you are trying to compute

$$\int \cot^3 x\,dx$$

and not

$$\int \cot x^3\, dx$$

right?

Hint: Try the substitution u=sin x.

3. Jul 17, 2010

### Chandasouk

Yes, Cot3x

4. Jul 18, 2010

### HallsofIvy

And is it
$$\frac{cot^3 (x)}{10}$$
or
$$cot^3\left(\frac{x}{10}\right)$$
?

Try writing $cot^3(x)$ as

$$\frac{sin^3(x)}{cos^3(x)}=$$$$\frac{sin^2(x)}{cos^3(x)}cos(x)=$$$$\frac{1- cos^2(x)}{cos^3(x)} sin(x)$$

and use the substitution u= cos(x).

5. Jul 18, 2010

### Chandasouk

It is $$\frac{cot^3 (x)}{10}$$

I actually have no idea how to use most of the math tags, so writing it out is hard.

Cot3x is $$\frac{Cos^3(x)}{Sin^3(x)}$$

I'll try distributing out a cosine factor and converting the rest to sign and see how it goes